
Let us assume that there are two identical conducting spheres M and N that have charges ${{q}_{m}}$ and ${{q}_{n}}$ respectively. Let a third similar neutral sphere, P , be brought in contact with M and then separated. Now sphere P is brought in contact with N and then it is made separated. Then what will be the final charge on sphere P?
$\begin{align}
& A.\dfrac{{{q}_{m}}+2{{q}_{n}}}{6} \\
& B.\dfrac{{{q}_{m}}+{{q}_{n}}}{4} \\
& C.{{q}_{m}}+\dfrac{{{q}_{n}}}{4} \\
& D.\dfrac{{{q}_{m}}+2{{q}_{n}}}{4} \\
\end{align}$
Answer
559.5k+ views
Hint: First of all find the charge acquired by the sphere P when kept in contact with the sphere M. initially the charge on the sphere P will be zero. Then calculate the charge acquired by the sphere P when kept in contact with the sphere N. this time the initial charge will become equal to the charge acquired by the sphere P when kept in contact with sphere M.
Complete step by step solution:
The charge on the sphere M has been given as ${{q}_{m}}$ and the charge on the sphere N has been given as ${{q}_{n}}$.
When the sphere P is brought in contact with the sphere M, then the charge on P will become,
${{q}_{p}}=\dfrac{0+{{q}_{m}}}{2}$
Initially the charge of the sphere P will be zero.
After making the two spheres to be in contact, the charge on the sphere P will become,
${{q}_{p}}=\dfrac{{{q}_{m}}}{2}$
Now the sphere P is brought in contact with N the amount charge will get varied in both the spheres. During this process initially the sphere P will be having a charge mentioned as,
${{q}_{p}}=\dfrac{{{q}_{m}}}{2}$
And initially the charge in the sphere N will be mentioned as ${{q}_{n}}$.
Therefore after making these sphere in contact, the charge on the sphere P will become,
${{{q}'}_{p}}=\dfrac{\dfrac{{{q}_{m}}}{2}+{{q}_{n}}}{2}$
Simplifying this will give,
${{{q}'}_{p}}=\dfrac{{{q}_{m}}+2{{q}_{n}}}{4}$
Therefore the final charge on the sphere will become,
${{{q}'}_{p}}=\dfrac{{{q}_{m}}+2{{q}_{n}}}{4}$
Hence the answer for the question has been calculated.
This answer has been given as option D.
Note:
Electric charge is defined as the basic physical characteristic of matter that will cause it to experience a force if it is placed in an electromagnetic field. There are two kinds of electric charges. One is classified as positive and the other is as negative.
Complete step by step solution:
The charge on the sphere M has been given as ${{q}_{m}}$ and the charge on the sphere N has been given as ${{q}_{n}}$.
When the sphere P is brought in contact with the sphere M, then the charge on P will become,
${{q}_{p}}=\dfrac{0+{{q}_{m}}}{2}$
Initially the charge of the sphere P will be zero.
After making the two spheres to be in contact, the charge on the sphere P will become,
${{q}_{p}}=\dfrac{{{q}_{m}}}{2}$
Now the sphere P is brought in contact with N the amount charge will get varied in both the spheres. During this process initially the sphere P will be having a charge mentioned as,
${{q}_{p}}=\dfrac{{{q}_{m}}}{2}$
And initially the charge in the sphere N will be mentioned as ${{q}_{n}}$.
Therefore after making these sphere in contact, the charge on the sphere P will become,
${{{q}'}_{p}}=\dfrac{\dfrac{{{q}_{m}}}{2}+{{q}_{n}}}{2}$
Simplifying this will give,
${{{q}'}_{p}}=\dfrac{{{q}_{m}}+2{{q}_{n}}}{4}$
Therefore the final charge on the sphere will become,
${{{q}'}_{p}}=\dfrac{{{q}_{m}}+2{{q}_{n}}}{4}$
Hence the answer for the question has been calculated.
This answer has been given as option D.
Note:
Electric charge is defined as the basic physical characteristic of matter that will cause it to experience a force if it is placed in an electromagnetic field. There are two kinds of electric charges. One is classified as positive and the other is as negative.
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